OF ARTS AND SCIENCES. 43 



times as abscissas. Next draw a similar curve wheu tlic tlicrmo cur- 

 rent is passing. Let the current run through a galvanometer, and ' 

 observe the deflection of the galvanometer every time a reading of the 

 thermometer is taken. This will give a third curve, giving the ther- 

 mo-electric current at any time, corresponding to any temperature of 

 the second curve. Finally, pass a current of known strength / from 

 an external source through the junction in the direction of the proper 

 thermo current, and get a fourth curve representing the fall of tem- 

 perature for this case. 



The equations of all these curves being known, we can find from the 

 two first the rate at which heat is absorbed by a thermo-electric cur- 

 rent of any strength given by the third curve. Let h = the heat 

 absorbed in unit time by this thermo-electric current of strength i. 

 From the first and fourth carves we can find the amount of heat H 

 absorbed in unit time by the battery current I. 



Now, if the heat H is merely the heat absorbed by the Peltier effect, 

 we have — as the heat of the Peltier effect is simply proportional to 

 the current strength : — 



h :H::i:L (4) 



But if, as I supposed, h was much greater than the heat absorbed by 

 the Peltier effect, this equation would not be satisfied. 



The thermo element used was of German silver and iron. The hot 

 junction was shaped in the form of a ring and placed in a small vessel 

 of mercury, the bulb of the thermometer being placed in the centre of 

 the ring. The first two curves, however, were identical, though Ger- 

 man silver and iron constitute one of the sti'ongest thermo elements. 

 The thermometer fell at exactly the same rate whether the current 

 was passing or not. 



We have not, however, considered the Thomson effect. But the 

 same reasoning used in the case of the Peltier effect applies also here. 



From equation (y) it was seen that the proper thermo-electric current 

 was proportional to 



[p-i — 0-2) d9 

 e' 

 Let 



*J fl' 



L 



(o-i — 0-2) de = s 



Then this current is proportional to 9 -|- *S'. That is, 9 -f- aS is 

 the whole heat absorbed in the circuit by unit current in unit time. 



